Rewrite the function by completing the square. $f(x)=x^{2}-2x+17$ $f(x)=(x+$
Solution: We want to complete $x^2{-2}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-2}}{2}\right)^2={1}$ to it: $x^2{-2}x+{1}=(x-1)^2$ In order to keep the expression equivalent, we add and subtract ${1}$, not forgetting the expression's constant term, $17$ : $\begin{aligned} f(x)&=x^2-2x+17 \\\\ &=x^2-2x+{1}+17-{1} \\\\ &=(x-1)^2+17-1 \\\\ &=(x-1)^2+16 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 1)^2 + 16$ This is equivalent to $f(x)=(x+{-1})^2+16$